Solve the differential equations (t - 1)3 ds/dt + 4(t

Gregory Jones 2021-12-31 Answered
Solve the differential equations \(\displaystyle{\left({t}-{1}\right)}{3}{d}\frac{{s}}{{\left.{d}{t}\right.}}+{4}{\left({t}-{1}\right)}{2}{s}={t}+{1},{t}{>}{1}\)

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Expert Answer

censoratojk
Answered 2022-01-01 Author has 4897 answers
Step 1
Given: \(\displaystyle{\left({t}-{1}\right)}^{{{3}}}{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{4}{\left({t}-{1}\right)}^{{{2}}}\ {s}={t}+{1};{t}{>}{1}\)
Step 2
Calculation:
\(\displaystyle{\left({t}-{1}\right)}^{{{3}}}{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{4}{\left({t}-{1}\right)}^{{{2}}}{s}={t}+{1};{t}{>}{1}\)
\(\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{\frac{{{4}{\left({t}-{1}\right)}^{{{2}}}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}{s}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}\)
\(\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{\frac{{{4}}}{{{\left({t}-{1}\right)}}}}\ {s}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}\)
Compare the differential equation with \(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{P}{y}={Q}\)
Here \(\displaystyle{P}={\frac{{{4}}}{{{\left({t}-{1}\right)}^{{{2}}}}}}\ \text{and}{Q}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}\)
Solution of given differential equation is \(\displaystyle{y}.{\left({I}.{F}\right)}=\int{Q}.{\left({I}.{F}\right)}{\left.{d}{x}\right.}+{c}\)
Now, \(\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}.{\left({t}-{1}\right)}^{{{4}}}{\left.{d}{t}\right.}+{c}\)
\(\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\left({t}+{1}\right)}.{\left({t}-{1}\right)}{\left.{d}{t}\right.}+{c}\)
\(\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\left({t}^{{{2}}}-{1}\right)}{\left.{d}{t}\right.}+{c}\)
\(\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}={\frac{{{t}^{{{3}}}}}{{{3}}}}-{t}+{c}\)
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godsrvnt0706
Answered 2022-01-02 Author has 394 answers
\(\displaystyle{\left({t}-{1}\right)}^{{{3}}}{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{4}{\left({t}-{1}\right)}^{{{2}}}\ {s}={t}+{1};{t}{>}{1}\)
\(\displaystyle{\left({t}-{1}\right)}^{{{3}}}{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{4}{\left({t}-{1}\right)}^{{{2}}}{s}={t}+{1};{t}{>}{1}\)
\(\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{\frac{{{4}{\left({t}-{1}\right)}^{{{2}}}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}{s}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}\)
\(\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}+{\frac{{{4}}}{{{\left({t}-{1}\right)}}}}\ {s}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{P}{y}={Q}\)
\(\displaystyle{P}={\frac{{{4}}}{{{\left({t}-{1}\right)}^{{{2}}}}}}\ \text{and}{Q}={\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}\)
\(\displaystyle{y}.{\left({I}.{F}\right)}=\int{Q}.{\left({I}.{F}\right)}{\left.{d}{x}\right.}+{c}\)
\(\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\frac{{{t}+{1}}}{{{\left({t}-{1}\right)}^{{{3}}}}}}.{\left({t}-{1}\right)}^{{{4}}}{\left.{d}{t}\right.}+{c}\)
\(\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\left({t}+{1}\right)}.{\left({t}-{1}\right)}{\left.{d}{t}\right.}+{c}\)
\(\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}=\int{\left({t}^{{{2}}}-{1}\right)}{\left.{d}{t}\right.}+{c}\)
\(\displaystyle{s}.{\left({t}-{1}\right)}^{{{4}}}={\frac{{{t}^{{{3}}}}}{{{3}}}}-{t}+{c}\)
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karton
Answered 2022-01-10 Author has 8454 answers

Solution:
\((t-1)^{3} \frac{ds}{dt}+4(t-1)^{2}s=t+1; t>1\)
\(\frac{ds}{dt}+\frac{4(t-1)^{2}}{(t-1)^{3}} s=\frac{t+1}{(t-1)^{3}}\)
\(\frac{ds}{dt}+\frac{4}{(t-1)}\ s=\frac{t+1}{(t-1)^{3}}\)
Compare the differential equation with \(\frac{dy}{dx}+Py=Q\)
\(P=\frac{4}{(t-1)^{2}}\ \text{and} Q=\frac{t+1}{(t-1)^{3}}\)
Solution of given differential equation is \(y. (I.F)=\int Q. (I.F)dx+c\)
\(s.(t-1)^{4}=\int \frac{t+1}{(t-1)^{3}}.(t-1)^{4}dt+c\)
\(s.(t-1)^{4}=\frac{t^{3}}{3}-t+c\)

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